# When 8.05 g of an unknown compound X was dissolved in 100. g of benzene (C6H6), the vapor pressure of the benzene decreased from 100.0 torr to 94.8 torr at 26 degrees C. What is the mole fraction and molar mass of X?

Feb 23, 2017

I got $\text{114.51 g/mol}$.

The reduction of vapor pressure is a colligative property. From Raoult's law, we have:

${P}_{j} = {\chi}_{j} {P}_{j}^{\text{*}}$,

where ${P}_{j}$ is the partial pressure above the solution from component $j$, ${P}_{j}^{\text{*}}$ is the pure vapor pressure of $j$ by itself, and ${\chi}_{j}$ is the mol fraction of $j$ in the solution.

(We consider $\boldsymbol{j}$ to be the solvent and $\boldsymbol{i}$ to be the solute.)

But mol fractions are always $\le 1$. Therefore, for a two-component solution, ${\chi}_{j} = 1 - {\chi}_{i}$. Thus:

${P}_{j} = \left(1 - {\chi}_{i}\right) {P}_{j}^{\text{*}}$

Since ${\chi}_{j} \le 1$, it follows that $1 - {\chi}_{i} \le 1$ as well, or that ${\chi}_{i} \ge 0$. Therefore, adding any solute into a solvent decreases its vapor pressure.

This reduction is calculated as follows. Define $\Delta {P}_{j} = {P}_{j} - {P}_{j}^{\text{*}}$ to get:

$\textcolor{g r e e n}{\Delta {P}_{j}} = {\chi}_{j} {P}_{j}^{\text{*" - P_j^"*}}$

$= \left({\chi}_{j} - 1\right) {P}_{j}^{\text{*}}$

$= \left(1 - {\chi}_{i} - 1\right) {P}_{j}^{\text{*}}$

$= \textcolor{g r e e n}{- {\chi}_{i} {P}_{j}^{\text{*}}}$

Since ${P}_{j}^{\text{*}}$ and ${\chi}_{i}$ are positive, the reduction in vapor pressure is of course negatively-signed. From this equation then, we can determine the mol fraction of what was dissolved in benzene.

Plug ${P}_{j} = \text{94.8 torr}$ and ${P}_{j}^{\text{*" = "100.00 torr}}$ in to get:

(94.8 - 100.00) "torr" = -chi_i ("100.00 torr")

$\implies \textcolor{b l u e}{{\chi}_{i} = 0.052}$

Now that we have the mol fraction of $X$, we can find the molar mass of $X$ simply by solving for the mols of $X$, and dividing the given mass by the mols of $X$.

$0.052 = \frac{{n}_{X}}{{n}_{X} + {n}_{\text{Ben}}}$

$\implies 0.052 {n}_{X} + 0.052 {n}_{\text{Ben}} = {n}_{X}$

$\implies \left(0.052 - 1\right) {n}_{X} = - 0.052 {n}_{\text{Ben}}$

$\implies {n}_{X} = 0.0549 {n}_{\text{Ben}}$

But we know the molar mass of benzene; it's $6 \times 12.011 + 6 \times 1.0079 = \text{78.1134 g/mol}$. So, we get the mols of benzene by its molar mass:

$\text{100.0 g Benzene" xx "1 mol"/"78.1134 g" = "1.2802 mols Benzene solvent}$

Therefore, the mols of $X$ are:

n_X = 0.0549("1.2802 mols") = "0.0703 mols X"

This gives a molar mass for $X$ as:

color(blue)(M_(m,X)) = "8.05 g X"/"0.0703 mols X" = color(blue)("114.51 g/mol")